A charge of +q is located at the origin, while an identical charge is located on the x axis at x = 0.45 m. A third charge of +4q is located on the x axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located?
I am just supposed to be using Coulombs law. I tried in relation to origin charge where the charge at .45 would be F=kq/.45^2 and in relation with charge 4q 4q/X^2. Then since q1=q2 the formula was kq/.45^2 = k4q/x^2. simplified that into x^2=4/.45^2. Then took the square root of the answer. it was like 4.444 totally wrong. I cannot figure out this problem at all please some guidance!
To solve this problem, let's consider the forces acting on the charge at the origin.
The electrostatic force between two charges q1 and q2 is given by Coulomb's Law:
F = (k * |q1| * |q2|) / r^2
where F is the force, k is the electrostatic constant (k = 9 x 10^9 N m^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between the charges.
In this case, let's assume the distance between the charge at the origin (+q) and the charge at x = 0.45 m (+q) is d.
The force between these two charges is given by:
F1 = (k * q * q) / d^2
Now we want to find the position (x) on the x-axis where a charge of +4q should be placed, such that the net force on the charge at the origin doubles.
Let's denote the position of the third charge (+4q) on the x-axis as x2.
The force between the charge at the origin (+q) and the charge at x2 (+4q) is given by:
F2 = (k * q * 4q) / (x2 - 0)^2
According to the problem, we want the net electrostatic force on the charge at the origin to double, so:
2 * F1 = F1 + F2
Substituting the expressions for F1 and F2, we get:
2 * [(k * q * q) / d^2] = [(k * q * q) / d^2] + [(k * q * 4q) / (x2 - 0)^2]
Simplifying this equation, we can cancel out some terms:
2/d^2 = 1/d^2 + 4/(x2^2)
Multiply both sides by d^2 to eliminate the denominators:
2 = 1 + 4(d^2 / x2^2)
Rearranging this equation, we get:
x2^2 = 4d^2
Taking the square root of both sides gives:
x2 = 2d
Therefore, the third charge (+4q) should be placed at a distance of 2d from the origin charge (+q) on the x-axis.
In this specific problem, the charge at the origin is located at the origin itself, so d = 0.45 m. Thus, the third charge should be placed at a distance of 2d = 2 * 0.45 m = 0.9 m from the origin on the x-axis.
To solve this problem, we can use Coulomb's law, which states that the electrostatic force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Let's begin by labeling the charges:
- The charge at the origin is +q.
- The charge at x = 0.45 m is also +q.
- The third charge is +4q.
We are asked to find the position of the third charge on the x-axis such that the net electrostatic force on the charge at the origin doubles, while its direction remains unchanged.
Step 1: Determine the initial force on the charge at the origin:
The electrostatic force between the charges at the origin and at x = 0.45 m can be calculated using Coulomb's law:
F₁ = k * q * q / (0.45 m)²
Step 2: Set up the equation for the desired force:
We are looking for the position of the third charge, so let's call it x. The force between the charges at the origin and at x will be:
F₂ = k * q * (4q) / (x)²
Step 3: Set up the equation for the desired force doubling:
Since we want the net electrostatic force to double, we can write:
2 * F₁ = F₂
2 * (k * q * q / (0.45 m)²) = k * q * (4q) / (x)²
Step 4: Solve the equation for x:
Now, we can rearrange the equation and solve for x:
2 * (0.45 m)² = (x)² / 4
Step 5: Calculate the location of the third charge:
Let's simplify the equation:
0.45 m = x / 2
Now, we can solve for x:
x = 0.45 m * 2
x = 0.9 m
Therefore, the third charge should be located at x = 0.9 m on the x-axis to double the net electrostatic force on the charge at the origin, while its direction remains unchanged.
To double the force at the origin, the +4q charge must have the same effect upon the charge as the +q charge at x=0.45 m
Making the distance x = 0.9 (twice that of the second charge) will result in the same force produced by the +4q particle, because of the the inverse square law.
You were OK with this equation
q/(0.45)^2 = 4q/(0.9)^2
Your mistake was algebra in the next step.
k*q/.45^2 = k*4q/x^2.
leads to
x^2 = 4*(0.45)^2
x = 0.9